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\(VT=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)
\(VT\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
Đặt cái ban đầu là P
Ta có: \(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta lại có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta có:
\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
thật vĩ đại Hung nguyen
Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)
\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)
Tương tự ta được
\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)
\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :
\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)
\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)
\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)
Ta có: \(x+y\ge2\sqrt{xy}\Rightarrow3xy\ge2\sqrt{xy}+1\Rightarrow3xy-2\sqrt{xy}-1\ge0\)
\(\Rightarrow\left(3\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)\ge0\Rightarrow\sqrt{xy}-1\ge0\) (do \(3\sqrt{xy}+1>0\) )
\(\Rightarrow\sqrt{xy}\ge1\Rightarrow xy\ge1\Rightarrow1-xy\le0\)
\(P=\dfrac{y\left(x+1\right)+x\left(y+1\right)}{xy\left(x+1\right)\left(y+1\right)}=\dfrac{2xy+x+y}{xy\left(xy+x+y+1\right)}\)
\(\Rightarrow P=\dfrac{2xy+3xy-1}{xy\left(xy+3xy\right)}=\dfrac{5xy-1}{4\left(xy\right)^2}=\dfrac{-4\left(xy\right)^2+5xy-1}{4\left(xy\right)^2}+1\)
\(\Rightarrow P=\dfrac{\left(1-xy\right)\left(4xy+1\right)}{4\left(xy\right)^2}+1\)
Do \(\left\{{}\begin{matrix}1-xy\le0\\4xy+1>0\\4\left(xy\right)^2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(1-xy\right)\left(4xy+1\right)}{4\left(xy\right)^2}\le0\)
\(\Rightarrow P\le0+1=1\Rightarrow P_{max}=1\) khi \(x=y=1\)
\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)
Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Rightarrow abc\ge1\)
\(BĐT\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)
Ta có \(abc\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}bc\ge\dfrac{1}{a}\\ab\ge\dfrac{1}{c}\\ac\ge\dfrac{1}{b}\end{matrix}\right.\Rightarrow bc+ac+ab\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)
1/
\(S=\dfrac{1}{x}+\dfrac{2^2}{y}\ge\dfrac{\left(1+2\right)^2}{x+y}=\dfrac{9}{1}=9\)
\(\Rightarrow S_{min}=9\) khi \(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{2}{y}\\x+y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
2/
Áp dụng BĐT: \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\Rightarrow x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Rightarrow\dfrac{\left(x+y\right)^2}{2}-3\left(x+y\right)\le x^2+y^2-3\left(x+y\right)=-4\)
\(\Rightarrow\dfrac{\left(x+y\right)^2}{2}-3\left(x+y\right)+4\le0\Leftrightarrow\left(x+y\right)^2-6\left(x+y\right)+8\le0\)
Đặt \(x+y=a\Rightarrow a^2-6a+8\le0\Rightarrow2\le a\le4\)
\(\Rightarrow2\le x+y\le4\)
\(\Rightarrow S\in\left[2;4\right]\)
1. Theo BĐT AM - GM, ta có:
\(\Sigma\dfrac{1}{\left(2x+y+z\right)^2}=\Sigma\dfrac{1}{\left\{\left(x+y\right)+\left(x+z\right)\right\}^2}\le\Sigma\dfrac{1}{4\left(x+y\right)\left(x+z\right)}\)
Do đó BĐT ban đầu sẽ đúng nếu ta C/m được
\(\Sigma\dfrac{1}{4\left(x+y\right)\left(x+z\right)}\le\dfrac{3}{16}\Leftrightarrow\dfrac{8}{3}\left(x+y+z\right)\le\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow\dfrac{8}{3}\left(x+y+z\right)\left(xy+yz+zx\right)\le\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(xy+yz+zx\right)\)
Nhưng điều này đúng vì \(xy+yz+zx\ge\sqrt[3]{x^2y^2z^2}=3\) và theo bổ đề bên trên. Từ đó ta có điều phải chứng minh. Dấu bằng xảy ra \(\Leftrightarrow a=b=c=1\)
( Còn bài 2 để suy nghĩ rồi tối đăng cho nha )
Để ý đẳng thức : \(\dfrac{xy}{\left(y-z\right)\left(z-x\right)}+\dfrac{yz}{\left(z-x\right)\left(x-y\right)}+\dfrac{xz}{\left(x-y\right)\left(y-z\right)}=\dfrac{xy\left(x-y\right)+yz\left(y-z\right)+xz\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=-\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=-1\)
Ta luôn có: \(\left(\dfrac{x}{y-z}+\dfrac{y}{z-x}+\dfrac{z}{x-y}\right)^2\ge0\) ;\(\forall x;y;z\)
\(\Leftrightarrow\dfrac{x^2}{\left(y-z\right)^2}+\dfrac{y^2}{\left(z-x\right)^2}+\dfrac{z^2}{\left(x-y\right)^2}\ge-2\sum\dfrac{xy}{\left(y-z\right)\left(z-x\right)}=2\)
(ĐPcm)
Dấu = xảy ra khi \(\dfrac{x}{y-z}+\dfrac{y}{z-x}+\dfrac{z}{x-y}=0\)
Thêm 1 ý tưởng đc buff từ cách trước :))
\(BDT\LeftrightarrowΣ\dfrac{x^2}{\left(y-z\right)^2}-2=\left(Σ\dfrac{x}{y-z}\right)^2-2Σ\dfrac{xy}{\left(y-z\right)\left(z-x\right)}-2\)
\(=\dfrac{\left(Σ\left(x^3-x^2y-x^2z+xyz\right)\right)^2}{\prod\left(x-y\right)^2}-2\dfrac{Σ\left(x^2y-x^2z\right)}{\prod\left(x-y\right)}-2\)
\(=\dfrac{\left(Σ\left(x^3-x^2y-x^2z+xyz\right)\right)^2}{\prod\left(x-y\right)^2}\ge0\)